1. CS 250, Discrete Structures, Fall 2015
Lecture 4.2: Relations
CS 250, Discrete Structures, Fall 2015
Nitesh Saxena
Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

2. Course Admin Mid-Term 2 Exam HW3
Solution will be posted soon
Should have the results this week
HW3
Was due just now
Solution will be posted
Should have the results soon
If you haven’t picked up your previous graded submissions, please do so from my office
1/17/2019

3. Roadmap PLEASE DO ATTEND THE LECTURES
Done with about 75% of the course
We are left with one more homework
We are left with one final exam
Cumulative – will cover everything
Focus more on material covered after the two mid-terms
Please try to do your best in the remainder of the time; I know you have been working hard
Please submit your homeworks on time
Still a lot of scope of improvement
PLEASE DO ATTEND THE LECTURES
1/17/2019

4. Outline Relation Examples and Definitions Matrix Representation
1/17/2019

5. Composing Relations Q: Suppose R defined on N by: xRy iff y = x 2
and S defined on N by: xSy iff y = x 3
What is the composition SR ?
1/17/2019

6. Composing Relations xRy iff y = x 2 xSy iff y = x 3
A: These are functions (squaring and cubing) so the composite SR is just the function composition (raising to the 6th power). xSRy iff y = x 6 (in this odd case SR = RS)
Q: Compose the following:
5 5
More detail on: xRSy iff y = x 6
1) xRx2
2) ySy3 so x2S(x2) 3
3) Composing (1) and (2) we get xRSx6
1/17/2019

7. Composing Relations
4 4 5
A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 4
1/17/2019

8. Composing Relations: Picture
4 4 5
A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 4
1/17/2019

9. Composing Relations: Picture
4 4 5
A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 4
1/17/2019

10. Composing Relations: Picture
4 4 5
A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 4
1/17/2019

11. Composing Relations: Picture
4 4 5
A: Draw all possible shortcuts. In our case, all shortcuts went through 1: 4
1/17/2019

12. Inverting Relations
Relational inversion amounts to just reversing all the tuples of a binary relation.
DEF: If R is a relation from A to B, then the relation R -1 from B to A defined by setting bR -1a if and only aRb.
Q: Suppose R defined from Z to N by: xRy iff y = x 2. What is the inverse R -1 ?
1/17/2019

13. Inverting Relations A: xRy iff y = x 2.
R is the square function so R -1 is square root: i.e. the union of the two square-root branches. I.e:
yR -1x iff y = x 2
or in terms of square root:
xR -1y iff y = ±x where x is non-negative
1/17/2019

14. Relations – matrix representation
Suppose we have a relation R on AxB, where A={1,2,3,4}, and B={u,v,w},
R={(1,u),(1,v),(2,w),(3,w),(4,u)}.
Then we can represent R as:
The labels on the outside are for clarity. It’s really the matrix in the middle that’s important. u v w 1 2 3 4
This is a |A| x |B| matrix whose entries indicate membership in R.
1/17/2019

15. Relations – matrix representation
Some things to think about.
Let R be a relation on a set A, and let MR be the matrix representation of R.
Then R is reflexive if, ______________.
All entries in MR are 1.
The \ diagonal of MR contains only 1s.
The first column of MR contains no 0s.
None of the above.
1/17/2019

16. Relations – matrix representation
Some things to think about.
Let R be a relation on a set A, and let MR be the matrix representation of R.
Then R is symmetric if, ______________.
All entries above the \ are 1.
The first and last columns of MR contain an equal # of 0s.
MR is visually symmetric about the \ diagonal.
None of the above.
1/17/2019

17. Relations – matrix representation
Suppose we have R1 and R2 defined on A: R1 u v w 1 R2 u v w 1 1
Then R1  R2 is the bitwise “or” of the entries (Join By):
MR1R2 = MR1 v MR2 1
Then R1  R2 is the bitwise “and” of the entries (Meet):
MR1R2 = MR1  MR2
1/17/2019

18. Relations – composition using matrices
Suppose we have R and S defined on A: R u v w 1 S u v w 1 1
Then SR corresponds to the boolean product
1/17/2019

19. Relations - A Theorem Theorem:
If R is a transitive relation, then Rn  R, n.
How to prove? What strategy or technique should we use?
1/17/2019

20. Typical way of proving subset.
Relations - A Theorem
If R is a transitive relation, then Rn  R, n.
Proof by induction on n.
Base case (n=1): R1  R because by definition, R1 = R.
Induction case: if R is transitive, then Rk  R.
Prove: if R is transitive, then Rk+1  R.
Typical way of proving subset.
We are trying to prove that Rk+1  R. To do this, we select an element of Rk+1 and show that it is also an element of R.
Let (a,b) be an element of Rk+1. Since Rk+1 = Rk  R, we know there is an x so that (a,x)  R and (x,b)  Rk.
By assumption at the induction step, since Rk  R, (x,b)  R.
But wait, if (a,x)  R, and (x,b)  R, and R is transitive, then (a,b)  R.
1/17/2019

21. Relations - Another Theorem
If R is a reflexive relation, then Rn is reflexive relation, n.
Whiteboard!
1/17/2019

22. N-ary Relations
So far, we were talking about binary relations – defined on two sets.
Can be generalized to N sets
Ex: R = {(a, b, c): a < b < c}, defined on set of integers – a 3-ary relation
Applications in databases
1/17/2019

23. Today’s Reading
Rosen 9.1 and 9.3
1/17/2019